Fig.~\ref{fig:ex15img} shows the feasible integer solutions and the convex hull of $X$. Adding the two dotted lines to the given inequalities suffices to describe the convex hull of $X$ without an integrality condition.

\begin{figure}[ht!]
\centering
\includegraphics[height=50mm]{ex15img.png}
\caption{Feasible solutions as dots. Convex hull of $X$ checkered. Original inequalities as lines. Remaining inequalities to define convex hull as dotted lines.}
\label{fig:ex15img}
\end{figure}

The horizontal dotted line is close to the basic feasible solution of the $LP$ that is equal to the intersection of the lines $-4x_1+6x_2 \leq 9$ and $2x_1+3x_2 \leq 10$. Therefore it makes sense to look for multipliers $u_1$ and $u_2$ that don't vanish, and eliminate the variable $x_1$. A simple solution would be $\vc{u} = (\frac{1}{2},1)$ but this places the resulting valid inequality at $x_2 \leq \frac{7}{3}$. In order to arrive at a stronger inequality, we use a higher denominator: $\vc{u} = (\frac{1}{6}, \frac{1}{3})$. This yields

\begin{align*}
\lfloor 2 \rfloor x_2 & \leq \lfloor \frac{29}{6} \rfloor \\
2 x_2 & \leq 4 \\
x_2 & \leq 2.
\end{align*}

We see that the other dotted line is close to the inequality $-4x_1+6x_2 \leq 9$. Using the other inequality will not help us because that would decrease the slope, and we need to increase it. Therefore we set $u_2 = 0$. A suitable value for $u_1$ turns out to be $\frac{1}{5}$:

\begin{align*}
\lfloor -\frac{4}{5} \rfloor x_1 + \lfloor \frac{6}{5} \rfloor x_2 & \leq  \lfloor \frac{9}{5} \rfloor \\
-x_1+x_2 & \leq 1.
\end{align*}